storing system time to a variable in cpp - c++

Question

I want to save the local time to a char variable. here is the code i have used. but it is saying
"cannot convert char * to char"
Here is my code:
#include <stdio.h>
#include <time.h>
struct tme
{
char intime;
}e;
void main( )
{
char timeStr [9];
_strtime( timeStr );
e.intime=timeStr;
printf( "The current time is %s \n", timeStr);
}
Thanx in advance.

Answer 1

That's simple, you have a char array timeStr of length 9 and trying to assign it to a char intime. There's type incompatibility. Think of it as char[] is never equal to char.
You could solve this as follows (but I don't know what you want to achieve):
struct tme
{
char* intime;
}e;
PS: MSDN states that (_strtime):
// Note: _strtime is deprecated; consider using _strtime_s instead

Answer 2

e.intime=timeStr;
timeStr is of type char [9]. It decays to a pointer pointing to first element during assignments or in a function call used as a parameter.
e.intime is of type char. char and char* aren't type compatible and the compiler is complaining you that. Instead you can do -
struct tme
{
char intime[10]; // +1 for the termination character to play safe
}e;
Now, strcpy can be used to copy the time to the member variable.
strcpy(e.intime, timeStr);
If it is C++, use std::string instead of raw arrays.

Answer 3

Some stages of refinement:
Stage 1: Fix your code.
struct tme {
char * intime; // You had a type mismatch
} e;
int main () { // Don't use void main()
char timeStr [9];
_strtime( timeStr );
e.intime=timeStr;
printf( "The current time is %s \n", timeStr);
}
There's a problem here: Your struct tme is relying on the external world to do everything for it, and to do so correctly. What if we want to reuse timeStr in main? What if you use this structure in a function other than main and set e.intime to a variable that goes out of scope?
Refinement: struct tme should own the time buffer.
struct tme {
char intime[9]; // Put the buffer here, not in main.
} e;
int main () {
_strtime( e.intime );
printf( "The current time is %s \n", e.intime);
}
We still have a problem here. That buffer can be modified by anyone, and the structure is just a passive receptacle.
Refinement: Hide the data and make the object active.
struct tme {
const char * set_time () { _strtime (intime); return intime; }
const char * get_time () const { return intime; }
private:
char intime[9];
};
int main () {
printf( "The current time is %s \n", e.set_time());
}